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Math Help - Deriving Mollweide's Formula

  1. #1
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    Deriving Mollweide's Formula

    For this post, A is alpha, B is beta, and Y is gamma, and represent the angles of the triangle. Little a, b, and c represent the sides opposite them, respectively. The form of the formula is:
    Code:
    a - b         sin(1/2[A - B]
    -----   =    -------------
      c            cos([1/2]Y)
    
    There is a step in the derivation that I don't understand. Here is the derivation from my solutions book, and so this post doesn't get too protracted, I'll stop at the point that I don't understand:
    
    a - b        a    b
    -----  =     -- - --
      c          c    c
    
    
    =  sin A     sin B
       ----   -  ----
       sin Y     sin Y
    
    
    = sin A - sin B
      -------------
          sin Y
    
    
    = 2sin( (A-B) / 2) cos( (A+B) / 2)
      ------------------------------    makes use of sum to product formula
               sin (2Y/2)
    
    
    = 2sin( (A-B) / 2) cos( (A+B) / 2)
      --------------------------------
               2sin(Y/2)cos(Y/2)
    
    
    = sin( (A-B) / 2) cos(PI/2 - Y/2)
      --------------------------------
               sin(Y/2)cos(Y/2)
    
    
    How does cos((A+B)/2) get turned into cos(PI/2 - Y/2)?
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  2. #2
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    Well you do know that \alpha  + \beta  + \gamma  = \pi \quad  \Rightarrow \quad \alpha  + \beta  = \pi  - \gamma.

    Now divide by 2.
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  3. #3
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    I didn't see that for some reason. Thanks.
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